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XYLENE POWER LTD.

FNR REACTIVITY

By Charles Rhodes, P.Eng., Ph.D.

REACTOR CORE:
In choosing the gap between the steel fuel tubes the issue is one of maximizing the average fuel density in the fuel bundle while not unduely decreasing the liquid sodium circulation and not imposing unreasonable cleanliness restrictions on the liquid sodium pool.

A related issue is that the maximum sodium temperature as liquid sodium passes through the fuel bundle needs to be limited to provide sufficient temperature safety margin at the upper end of the liquid sodium operating temperature range.

Based on all of these issues the center to center spacing between the square lattice fuel tubes was chosen to be:
(5 / 8) inch = 0.625 inch.

This dimensional choice sets the smallest initial intertube gap in the assembly at (1 / 8) inch, so filtering should be used to eliminate particulates larger than (1 / 32) inch in longest dimension.

Each fuel tube position has associated with it a reactor top surface area of about 1 tube per [0.625^2 inch^2]
= 1 tube / .3906 inch^2

The cross sectional area initially occupied by each fuel tube is:
Pi (.25 inch)^2 = 0.1963 inch^2

Thus the remaining cross sectional area per core fuel tube initially available for natural convection liquid sodium coolant flow is:
0.3906 inch^2 - 0.1963 inch^2 = 0.1943 inch^2
= 0.1943 inch^2 X (.0254 m / inch)^2
= 1.2535 X 10^-4 m^2

Note that initially the cross sectional area of a fuel tube OD and the cross sectional area of a flow channel are almost equal.

Note that a scale diagram shows that initially the flow channel wall area is almost equal to the OD wall area of a fuel tube.

Hence the initial effective cross sectional area for natural circulation through the reactor is:
1.2535 X 10^-4 m^2 / core fuel tube X 544 tubes / bundle X 532 core bundles = 36.2772 m^2

If the fuel tube radius linearly swells by 10% the cross sectional area occupied by each fuel tube becomes:
1.21 X 0.1963 inch^2 = 0.237523 inch^2

Thus the remaining cross sectional area per core fuel tube available for natural convection liquid sodium coolant flow after 10% linear tube swelling is:
0.3906 inch^2 - 0.237523 inch^ = 0.1531 inch^2
= 0.1531 inch^2 X (.0254 m / inch)^2
= 9.8759 X 10^-5 m^2

Fuel tube swelling also increases viscous effects that further reduce primary liqud sodium circulation.

The length of the reactor core perimeter is about:
Pi X 10.4 m = 32.67 m

Hence, prior to fuel tube swelling, the thickness of the radially horizontally moving liquid sodium layer at the active region perimeter is about:
(30.4089 m^2) / (32.67 m = .9307 m.
 

REACTIVITY REQUIREMENT
In order to sustain the nuclear chain reaction in the reactor core while controlling the reaction by thermal expansion the mean fast neutron travel along the Z axis between successive Pu-239 fissions must be greater than the core zone height. On the web page_________it was determined that a practical core zone height is about 0.35 m.

The neutron path between successive fissions is inversely proportional to the average fissionable fuel density in the reactor. Hence a major issue in FNR design is maximizing the ratio of metallic core fuel volume to total reactor core volume while maintaining adequate cooling and while minimizing unwanted neutron absorption.

In order for an FNR to run there must be a self sustaining chain reaction. This requirement imposes a constraint on the average atomic density of Pu-239 in the reactor core as compared to the average atomic density of U-238, Na-23, Fe, Cr and probability of neutron leakage out the surface of the active core zone.

In order to control the nuclear reaction full insertion of the control rod must render a fuel bundle subcritical under all conditions. Fully removing a control rod should make the reactor run provided that other 8 adjacent contol rods are also removed. In order to conserve neutrons the control rods should be made out of blanket rod material, so that the neutrons that the control rods absorb usefully convert U-238 into Pu-239.

Each core zone of a fuel bundle has volume Vcz of: Vcz = 0.400 m X 0.400 m X 0.35 m
= 0.0560 m^3

Within core zone volume Vcz are uranium-plutonium-zirconium fuel rods with a uranium-plutonium-zirconium volume of:
544 X Pi X (8.08 X 10^-3 m / 2)^2 X 0.35 m = .009762 m^3

within volumeVcz are steel fuel tubes with a steel volume of:
544 X 0.35 m X Pi X [(.25 inch)^2 - (0.185 inch)^2] X (.0254 m / inch)^2 = .0109115512 m^3

Within volume Vcz are 4 girders with a combined steel volume of:
4 X 0.35 m X 0.25 inch^2 X 5 X (.0254 m / inch)^2 = 0.00112903 m^3

Within volume Vcz are 4 girders with a combined steel volume of:
4 X 0.35 m X 0.25 inch^2 X 3 X (.0254 m / inch)^2 = 0.000677418 m^3

Within volume Vcz are 4 shroud walls with a combined steel volume of:
4 X 0.35 m X (1 / 16) inch X 15 inch X (.0254 m / inch)^2 = 0.00084677 m^3

Within volume Vcz are 4 control bundle walls with a combined steel volume of:
4 X 0.35 m X (1 / 16) inch X 10 inch X (.0254 m / inch)^2 = 0.0005645 m^3

Thus the steel volume Vi within Vcz is given by:
Vi = .0109115512 m^3 + 0.00112903 m^3 + 0.000677418 m^3 + 0.00084677 m^3 + 0.0005645 m^3
= 0.014129 m^3

The remaining core region volume consisting of:
Vs = 0.0560 m^3 - 0.014129 m^3 - .009762 m^3
= 0.032109 m^3
is filled with liquid sodium.

Define:
Ns = average number of sodium atoms / unit volume in reactor core zone
Ni = average number of iron atoms per unit volume in reactor core zone
Nc = average number of chromium atoms per unit volume in ractor core zone
Np = average number of plutonium atoms per unit volume in the reactor core zone with control bundle inserted at start
Npw = averge number of plutonium atoms per unit volume in reactor core zone with contro bundle withdrawn
Nz = average number of zirconium atoms per unit volume in the reactor core zone
Nu = average number of U-238 atoms / unit volume in reactor core zone with control bundle inserted
Nuw = average number of U-238 atoms / unit volume in reactor core zone with control bundle wiothdrawn
Nf = average number of fission product atoms / unit volume in reactor core at end of fuel bundle life with control bundle inserted
Rhos = mass density of liquid sodium = .927 gm / cm^3
Rhoi = mass density of iron = 7.874 gm / cm^3
Rhou = mass density of U-238 = 19.1 gm / cm^3
Rhop = mass density of Pu-239 = 19.8 gm / cm^3
Rhoc = average mass density of control rod slug material Aws = atomic weight of sodium = 23
Awi = atomic weight of iron = 55.845
Awc = atomic weight of Chromium = __________
Awu = atomic weight of uranium = 238
Awz = atomic weight of zirconium = 91.22
Awc = atomic weight of control rod atoms Av = Avogadro's Number

OK TO HERE

At the end of a fuel bundle life as set by fuel tube material swelling:
Nf = 2 (0.59 Np)
= 2 (0.59) (Np / Nu) Nu
= 2 (0.59) (2 / 7) Nu
= 0.33714 Nu

Recall that:
Mass Mu of U-238 in each core fuel rod is:
Mu = .7 (0.28725 kg / core rod)
= 0.201075 kg

Mass of Pu in each core fuel rod is:
Mp = 0.2 (0.28725 kg / core rod)
= 0.05745 kg

Mass Mz of Zr in each core fuel rod is:
Mz = 0.1 (0.28725 kg / core rod kg)
= 0.028725 kg / core rod

Mc = control rod mass in a core zone consisting of blanket rod material

Mc = (Pi (3.0 inch)^2 X 0.35 m) X (.0254 m / inch)^2 X (15.884 X 10^3 kg / m^3)
= 101.41 kg

Nuc = Mc X 0.9 X (1 mole / .238 kg) X Av / 0.056 m^3

Ns = (0.0369377 m^3) X (927 kg / m^3) X (1 mole / .023 kg) X Av] / (0.056 m^3)

Nu = [456 core rods X 0.201075 kg / core rod X (1 mole / .238 kg) X Av] / (0.056 m^3)
= [(456 core rods X 0.201075 kg / core rod) X (1 mole / .238 kg) X 6.023 X 10^23 atoms / mole] / (0.056 m^3)
= 41.435 X 10^26 U atoms / m^3

Hence:
(Nuc / Nu) = Mc X 0.9 X (1 mole / .238 kg) X Av / 0.056 m^3
/ [(456 core rods X 0.201075 kg / core rod) X (1 mole / .238 kg) X Av] / (0.056 m^3)
 
= [101.41 kg X 0.9] / [(456 core rods X 0.201075 kg / core rod)]
 
= 0.9954

(Ns / Nu) = {[ 0.0369377 m^3 X (927 kg / m^3) X (1 mole / .023 kg) X Av] / (0.056 m^3)}
/ [456 core rods X 0.201075 kg / core rod X (1 mole / .238 kg) X Av] / (0.056 m^3)
 
= {[ 0.0369377 m^3 X (927 kg / m^3) X (1 mole / .023 kg)]}
/ [456 core rods X 0.201075 kg / core rod X (1 mole / .238 kg)]
= 3.86434

Similarly:

Ni = [ 0.0108787 m^3 X Rhoi X (1 mole / Awi) X Av] / (0.056 m^3)

Hence:
(Ni / Nu) = {[ 0.0108787 m^3 X Rhoi X (1 mole / Awi) X Av] / (0.056 m^3)}
/ [456 core rods X 0.201075 kg / core rod X (1 mole / .238 kg) X Av] / (0.056 m^3)
 
= {[ 0.0108787 m^3 X 7874 kg / m^3 X (1 mole /0.055845 kg)]}
/ [456 core rods X 0.201075 kg / core rod X (1 mole / .238 kg)])
 
= 3.9815

Nz = [456 core rods X (Mz / core rod) X (1 mole / Awz) X Av] / (0.06 m^3)

Hence:
(Nz / Nu) = {[456 core rods X (Mz / core rod) X (1 mole / Awz) X Av] / (0.056 m^3)}
/ [456 core rods X 0.201075 kg / core rod X (1 mole / .238 kg) X Av] / (0.056 m^3)
 
{[(Mz / core rod) X (1 mole / Awz)]}
/ [0.201075 kg / core rod X (1 mole / .238 kg)]
 
{[(0.028725 kg / core rod) X (1 mole / .09122 kg))]} / [0.201075 kg / core rod X (1 mole / .238 kg)]
= 0.372725
 

Np / Nu = (.2 / .7) = 0.2857

Re-define:
Np = number of fissionable plutonium atoms per unit volume
Gu = number of neutrons emitted per average U-235 atomic fission = 2.6
Gp = number of neutrons emitted per average Pu-239 atomic fission = 3.1
Ge = number of neutrons emitted per average atomic fission that are absorbed by breeding
Data from Kaye & Laby for a FNR core:
Sigmaas = fast neutron absorption cross section of sodium = 0.0014 b
Sigmass = fast neutron scatter cross section of sodium = 2.62 b
Sigmaai = fast neutron absorption cross section of iron = .0086 b
Sigmasi = fast neutron scatter cross section of iron = 4.6 b
Sigmaaf = fast neutron absorption cross section of fission products = ____ b
Sigmaau = fast neutron absorption cross section of U-238 = 0.25 b
Sigmasu = fast neutron scatter cross section of U-238 = 9.4 b
Sigmaap = fast neutron absorption cross section of plutonium-239 = 0.040 b
Sigmaaz = fast neutron absorption cross section of zirconium = 0.0066 b Sigmafp = fast neutron fission cross section of plutonium-239 = 1.70 b
Sigmafu = fast neutron fission cross section of uranium-238 = 0.041 b
 

WITH NO CONTROL ROD PRESENT CONSIDER THE DISTANCE Lsc TRAVELLED BY A NEUTRON IN A CORE ZONE BETWEEN SUCCESSIVE SCATTERS:
Lsc = 1 / [Ns Sigmass + Ni Sigmasi + Nu Sigmasu]
= 1 / {Nu [(Ns / Nu) Sigmass + (Ni / Nu) Sigmasi + (Nu / Nu) Sigmasu]}
= 10^28 (b / m^2) / {(41.435 X 10^26 U atoms / m^3) [(3.86434) (2.62 b) + (3.9815)(4.6 b) + (1) (9.4 b)]
= 100 m / {41.435 [10.12 + 18.3149 + 9.4]}
= 100 m / {41.435 [37.835]}
= 0.063787 m
Note that as Lsc increases when Ns, Ni, Nu decrease. Hence Lsc increases with increasing temperature.
 

WITH NO CONTROL ROD PRESENT CONSIDER THE DISTANCE Lac TRAVELLED BY A NEUTRON IN THE REACTOR CORE ZONE BEFORE ABSORPTION:
Lac = 1 / [Ns Sigmas + Ni Sigmaai + Nu Sigmaau + Nu Sigmafu + Np Sigmaap + Np Sigmafp]
= 1 / {Nu [(Ns / Nu) Sigmaas + (Ni / Nu) Sigmaai + (Nu / Nu) Sigmaau + (Nu / Nu) Sigmafu + (Np / Nu) (Sigmaap + Sigmafp)]}
= 10^28 (b / m^2) / {(41.435 X 10^26 U atoms / m^3) [(3.86434) (.0014 b) + (3.9815)(.0086 b) + .25 b + .041 b + (0.2857)(.040 + 1.70)]
= 100 m / {41.435 [.00541 + .03424 + .25 + .041 + .497118]}
= 100 m / {41.435 [.82776]}
= 2.915 m
Note that Lac increases when Ns, Ni, Nu, Np decrease. Hence Lac increases with increasing temperature.
 

Hence the number of neutron scatters before absorption is:
(Lac / Lsc) = 2.915 m / 0.063787 m
= 45.71

Since neutron scattering takes place in a 3 dimensional random walk the distance travelled by a neutron along a single axis before absorption is:
[(La / Ls)^0.5 X (Ls / 3^0.5)]
= [(La Ls) / 3]^0.5 = [(2.915 m X 0.063787 m) / 3]^0.5
= .2489 m

Note that diffusion distance [(La Ls) / 3]^0.5 increases with increasing temperature.

Note that by comparison the core zone height is 0.35 m and the adjacent blanket thickness is 1.20 m. The diffusion distance is more than half the core zone height. Hence about half of the fission neutrons diffuse out of the core zone into the blanket zone and do not contribute to maintenance of the chain reaction. An increase in temperature decreases Ns, Ni, Nu, Np which increases:
[(La Ls) / 3]^0.5 which increases the fraction Fp of neutrons diffusing out of the core zone and hence not supporting the chain reaction. Hence the chain reaction will stop at a sufficiently high temperature.
 

OK TO HERE FIND Lsb AND Lab USE DEDICATED WEB PAGE FOR NEXT SECTION

CONSIDER THE CASE OF THE CONTROL ROD FULLY WITHDRAWN:
With respect to a flux neutrons in reactor the condition for criticality is:
(- Ns Sigmaas - Ni Sigmaai - Nu Sigmaau - Nz Sigmaaz - Np Sigmaap - Nf Sigmaaf + Gp Fp Np Sigmafp) > 0
where:
Gp = 3.1 neutrons / fission
Fp ~ 0.8 = Fraction of fission neutrons that remain in core region to potentially participate in subsequent fusion reactions
giving:
[(Gp Fp Np Sigmafp)- (Np Sigmaap)] > Ns Sigmaas + Ni Sigmaai + Nu Sigmaau + Nz Sigmaaz + Nf Sigmaaf or
Np [(Gp Fp Sigmafp) -(Sigmaap)] > Ns Sigmaas + Ni Sigmaai + Nu Sigmaau + Nz Sigmaaz + Nf Sigmaaf or
(Np / Nu) >
[(Ns / Nu) Sigmaas + (Ni/ Nu) Sigmaai + Sigmaau + (Nz / Nu) Sigmaaz + (Nf / Nu) Sigmaaf)] / [(Gp Fp Sigmafp) - Sigmaap)]

As indicated at FNR FUEL TUBE WEAR at the end of the fuel tube's useful life when the number of plutonium atom fissions equals 0.59 of the number of plutonium atoms initially present:
(Nf / Nu) = 2 (0.59 Np / Np) (Np / Nu)
= 2 (0.59) (.2857) = 0.3371.

Then numerical substitution gives:
(Np / Nu)
> [(Ns / Nu) Sigmaas + (Ni / Nu) Sigmaai + Sigmaau + (Nz / Nu) Sigmaaz + (Nf / Nu) Sigmaaf] / [(Gp Fp Sigmafp) - Sigmaap]
or
(Np / Nu) > [3.72135 (.0014 b) + 4.4816 (.0086 b) + .25 b + 0.372725 (.0066 b) + 0.3371 (Sigmaaf)] / [(3.1 Fp X 1.7 b) - .40 b]
or
[0.00521 b + .03854 b + .25 b + .00246 b + .3371 (Sigmaaf)] / [(Fp 5.27 b) - 0.40 b]
or
(Np / Nu) > [.29621 b + .3371 (Sigmaaf)] / [(Fp 5.27 b) - 0.40 b]
or
[(Fp 5.27 b) - 0.40 b] > [.29621 b + .3371 (Sigmaaf)] / (Np / Nu)
or
[(Fp 5.27 b) - 0.40 b] > [.29621 b + .3371 (Sigmaaf)] / (.2857)
or
[(Fp 5.27 b) - 0.40 b] > [1.03678 b + 1.1799 (Sigmaaf)]
or
[(Fp 5.27 b)] > [1.43678 b + 1.1799 (Sigmaaf)]
or
Fp > 0.2726 + 0.2239 (Sigmaaf / b)

This is the condition that must be satisfied for the chain reaction to run when the control rod is fully withdrawn. At the beginning of the fuel life the 2nd term is zero.
 

NOW CONSIDER THE CASE OF THE CONTROL ROD FULLY INSERTED:
With respect to a flux neutrons in reactor the condition for ensuring non-criticality is:
(Np / Nu)
< [(Ns / Nu) Sigmaas + (Ni / Nu) Sigmaai + Sigmaau + (Nz / Nu) Sigmaaz + (Nf / Nu) Sigmaaf + (Nuc / Nu) Sigmaau]
/ [(Gp Fp Sigmafp) - Sigmaap]
 
or
(Np / Nu) < [.29621 b + .3371 (Sigmaaf) + (1.3299) (.25 b)] / [(Fp 5.27 b) - 0.40 b]
or
(Np / Nu) < [.628685 b + .3371 (Sigmaaf)] / [(Fp 5.27 b) - 0.40 b]
or
[(Fp 5.27 b) - 0.40 b] < [.628685 b + .3371 (Sigmaaf)] / (Np / Nu)

This is the condition that must be satisfied for the chain reaction to stop when the control rod is fully inserted. Note that at the beginning of the fuel life the 2nd term is zero.

For compliance with this equation at the beginning of the fuel life:
[(Fp 5.27 b) - 0.40 b] < [.628685 b] / (Np / Nu)
or
R giving:
[(Fp 5.27 b) - 0.40 b] < [.628685 b] / 0.2857
or
Fp 5.27 b < 2.600 b
or
Fp < 0.4934
 

SUMMARY:
Reactor operation is highly dependent on the correct choice of core region height to give the desired value of Fp.

0.2726 + 0.2239 (Sigmaaf / b) < Fp < 0.4934

For safe shutdown and to realize operationinthe face of accumulated fission products we need to choose Fp ~ 0.45.

We must solve the diffusion equationto find the diffusion fluxes of neutrons from the core regioninto the adjacent blanket regions to accurately determine Fp in terms of the core rod length Lc.

If Fp is less than 0.27 the reactor will never run. If Fp is larger than 0.49 the chain reactin cannot be stopped by inserion of the contemplated U-238 control rod.

To find Fp we must solve the equation for the diffusion flux of neutronsinto the blanket regions above and below the core region.

Once fission products start to accumulate the condition for operation until the end of the fuel tube life is:
0.2726 + 0.2239 (Sigmaaf / b) < Fp
For Fp = 0.45: 0.2726 + 0.2239 (Sigmaaf / b) < 0.45
or
(Sigmaaf / b) < 0.794
or
(Sigmaaf) < 0.794 b

Thus to make maximum use of the fuel tube material the fast neutron cross section of the fission products must be less than 0.794 b. Based on the Kaye & Laby data it appears for fast neutrons the fission product cross sections are much smaller than 0.794 barnes so fuel bundle working life is determined by fuel tube swelling rather than by fission product accumulation.
 

SAFETY REQUIREMENT:
Note that for safety there must be absolute certainty that the chain reaction can be shut down, so Fp must be significantly less than 0.49. A safe initial choice is Fp = 0.45
 

THERMAL POWER CONTROL:
In a liquid sodium cooled FNR the primary means of thermal power control is thermal expansion which changes the fraction of fission neutrons that diffuse out of the core zone and hence regulates the chain reaction.
 

FUEL BUNDLE REACTIVITY:
The fuel bundles are engineered such that all core zones are sub-critical with the control rods fully inserted. Ideally if a single core fuel bundle in an array of core fuel bundles with their control rods inserted has its control rod withdrawn that fuel bundle should remain subcritical. This feature is desirable to prevent problems if a single control rod jams in a withdrawn position.

As the reactor core zone exceeds its design operating temperature its thermal expansion should cause the core zone to drop below criticality.

CORE ZONE HEIGHT:
On average each U-235 fission produces 2.6 neutrons and each Pu-239 fission produces 3.1 neutrons. In order for criticality to be maintined one of these three neutrons must be captured by a plutonium atom. That capture must happen before the neutron leaves the core zone.

Once a neutron leaves the core zone the probability of it being captured within the adhacent blanket zone must be vey high.

In a true breeder started with U-235 there is very little margin for neutron loss. Plutonium is a much more practical start fuel.

CRITICALITY MAINTENANCE:
Maintenance of criticality in the core zone requires a minimum core zone size together with a minimum plutonium density within the core zone. From each fission the chain reaction must use at least one neutron to sustain the chain reaction and will likely lose an additional 0.2 neutrons in the core. Hence roughly speeking, if La is the average total neutron travel distance from emission to fission:

However, in travelling distance La the neutron goes through about 50 scattering events so its linear travel distance is not that large.

BREEDING CONDITION:
The probability of surplus neutron capture by the surrounding U-238 needs to be at least 95%.

Let Lb = thickness of the blanket surrounding the reactor core.

The blanket is composed of fuel rods that do not contain fissionable material. Then the required minimum blanket thickness Lb is given by:
Lb = 3 /(Sigmaab Nub)
where Nub = average uranium atomic density in the blanket.

Nub =
[(3 X 546) blanket rods X (0.3112 kg / blanket rod) X (238 kg uranium / 260 kg blanket rod) X (6.023 X 10^23 atoms uranium / 0.238 kg uranium) / 0.2400 m^3
= 4.9201 X 10^27 atoms / m^3

Data from the UK Nuclear Data Library indicates that in the blanket the transport cross section is:
Sigmaab = 10.3 b. Hence the minimum required blanket thickness Lb is given by:
Lb = 3 /(Sigmaab Nu)
= 3 / {10.3 X 10^-28 m^2 X 4.9201 X 10^27 atoms / m^3}
= 0.592 m

I have provided for a blanket that is 1.2 m thick to ensure completeness of neutron absorption by U-238.
 

OPERATIONAL NOTE:
The Pu-239 concentration in the breeding blanket may be less than 2% by weight whereas in the core the Pu-239 concentration must be about 20% by weight. The Pu-239 concentration in the breeding blanket is gradually bred up to 2%. Then blanket rod reprocessing is used to remove almost 90% of the uranium and zirconium from the breeding blanket material to leave behind 20% Pu-239 material which is suitable as fuel for the reactor core.
 

PLUTONIUM DOUBLING TIME:
An issue of great importance in large scale implementation of FNRs is the FNR run time required for one FNR to breed enough excess Pu-239 to allow startup of another identical FNR. This time may be calculated using the approximation that each plutonium atom fission releases of 3.1 neutrons of which 2.5 neutrons are required for sustaining reactor operation leaving 0.6 neutrons for breeding extra Pu-239. Thus one atom of Pu-239 has to fission to form 0.6 atoms of extra Pu-239.

Hence the plutonium doubling time, which the time required to double the available amount of plutonium via breeding within the FNR is:
(10 / 6) X (time to consume the initial Pu supply)
= [(10 / 6) X (1 cycle time )] / 0.59
= 2.825 (1 fuel cycle time)
= 2.825 X (47.78 years)
= 136 years

This is the time required for one FNR to form enough excess Pu to allow starting another FNR. Clearly this doubling time is too long to enable rapid deployment of FNRs.

With large scale implementation of FNRs the available supply of plutonium and trans uranium actinides will soon be exhausted. Hence the issue of the Pu-239 doubling time physically constrains the rate of growth of the FNR fleet.

Thus FNRs are viable for disposing of transuranium actinides but due to the Pu-239 doubling time will not in the near future provide enough power capacity for complete displacement of fossil fuels.
 

This web page last partially updated January 29, 2017

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